What is (2x^2+3x-5)(3-5x)>= 0 ?

The solution to (2x^2+3x-5)(3-5x)>= 0 is x<=-5/2 \lor 3/5 <= x<= 1

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(2x^2+3x-5)(3-5x)>= 0

(2 x^{2} + 3 x - 5) (3 - 5 x) \geq 0

x \leq - \frac{5}{2} or \frac{3}{5} \leq x \leq 1

Interval Notation

(- \infty, - \frac{5}{2}] \cup [\frac{3}{5}, 1]

Decimal Notation

x \leq - 2.5 or 0.6 \leq x \leq 1

Solution steps

(2 x^{2} + 3 x - 5) (3 - 5 x) \geq 0

Factor

(2 x^{2} + 3 x - 5) (3 - 5 x) : - (x - 1) (2 x + 5) (5 x - 3)

- (x - 1) (2 x + 5) (5 x - 3) \geq 0

Multiply both sides by

- 1

(reverse the inequality)

(- (x - 1) (2 x + 5) (5 x - 3)) (- 1) \leq 0 \cdot (- 1)

Simplify

(x - 1) (2 x + 5) (5 x - 3) \leq 0

Identify the intervals

x \leq - \frac{5}{2} or \frac{3}{5} \leq x \leq 1

x^{6} - 9 x^{3} + 8 \leq 0

\frac{( - x ^{2} - 1 )}{( 9 - 2 x )} \leq 0

\frac{( 1 + 3 x ^{2} )}{( 2 x ^{2} - 21 x + 40 )} < 0

\frac{x ^{2} + 2 x - 3}{x ^{2} + 1} < 0

\frac{( x - 2 ) ^{2} ( x + 1 )}{( x - 3 )} > 0

What is (2x^2+3x-5)(3-5x)>= 0 ?
The solution to (2x^2+3x-5)(3-5x)>= 0 is x<=-5/2 \lor 3/5 <= x<= 1